# Characterize rings $R$, such that the countable product $P=R^N$ has the property that every finitely generated submodule of $P$ is free

What are the rings whose countable power has the property that every finitely generated submodule is free?

• Are your rings commutative with unity? Commented Jun 12 at 23:05
• @Alex Kruckman I would start with a commutative rings with unity. Once that becomes clarified, one can look at the more general case(s)
Commented Jun 15 at 3:46

I think your condition is equivalent to being a semifir (but see the remark below). See Chapter 2 of Paul Cohn, "Free ideal rings and localization in general rings". A commutative semifir is a Bézout domain, so in the commutative case your condition is equivalent to being a Bézout domain. This is more general than a PID.

For a semifir, you also need the ranks of finitely generated free modules to be well defined, so your condition might be a bit more general. Of course, in the commutative case, this is automatic.

• Great! Interestingly (or strangely), I have never seen this book of Cohn. Worth checking it out. Some people use tensoring to define rank, Another way is to imitate vector spaces.
Commented Jun 16 at 22:12
• Thanks. I will check Cohn's book, not sure I have it.
Commented Jun 16 at 22:24

Assuming rings are commutative with unity, these are exactly the Bézout domains (and the zero ring).

Let $$R$$ be a non-zero ring. The following are equivalent:

1. $$R$$ is a Bézout domain (a domain in which every f.g. ideal is principal).
2. Every f.g. torsion-free module is free.
3. Every f.g. submodule of $$R^\mathbb{N}$$ is free.
4. Every f.g. ideal is a free module.

$$1\implies 2$$ is the non-trivial implication. See Pete Clark's answer here. It also follows from Lemma 15.22.7 and Lemma 15.124.8 on the Stacks project.

$$2\implies 3$$ since $$R^\mathbb{N}$$ is torsion-free (so every submodule is torsion-free).

$$3\implies 4$$ since every ideal is a submodule of $$R$$, and hence a submodule of $$R^\mathbb{N}$$.

For $$4\implies 1$$, suppose $$R$$ is not a Bézout domain. Then $$R$$ has zero-divisors or a f.g. non-principal ideal. If $$a\in R$$ is a zero-divisor, then $$(a)$$ is a finitely generated ideal which is not a free module because it has torsion. If $$I\subseteq R$$ is a f.g. non-principal ideal, then $$I$$ is not a free module. Indeed, if $$a,b\in I$$ are non-zero, then $$ba-ab=0$$, so $$I$$ has rank at most $$1$$, but it is not principal, hence not free.

As you can see, there is nothing at all special about $$R^{\mathbb{N}}$$ being used in the proof. You can replace $$R^{\mathbb{N}}$$ with any other non-zero torsion-free module, and the equivalence holds.

• @DaveBenson Yes, thanks. Fixed. Commented Jun 16 at 12:05
• Interesting, we can then simply require the condition to be applied to R.